In mathematics, the disintegration theorem is a result in measure theory and probability theory. It rigorously defines the idea of a non-trivial "restriction" of a measure to a measure zero subset of the measure space in question. It is related to the existence of conditional probability measures. In a sense, "disintegration" is the opposite process to the construction of a product measure.

Consider the unit square ${\displaystyle S=[0,1]\times [0,1]}$ in the Euclidean plane ${\displaystyle \mathbb {R} ^{2}}$. Consider the probability measure ${\displaystyle \mu }$ defined on ${\displaystyle S}$ by the restriction of two-dimensional Lebesgue measure ${\displaystyle \lambda ^{2}}$ to ${\displaystyle S}$. That is, the probability of an event ${\displaystyle E\subseteq S}$ is simply the area of ${\displaystyle E}$. We assume ${\displaystyle E}$ is a measurable subset of ${\displaystyle S}$.

Consider a one-dimensional subset of ${\displaystyle S}$ such as the line segment ${\displaystyle L_{x}=\{x\}\times [0,1]}$. ${\displaystyle L_{x}}$ has ${\displaystyle \mu }$-measure zero; every subset of ${\displaystyle L_{x}}$ is a ${\displaystyle \mu }$-null set; since the Lebesgue measure space is a complete measure space,

$${\displaystyle E\subseteq L_{x}\implies \mu (E)=0.}$$

While true, this is somewhat unsatisfying. It would be nice to say that ${\displaystyle \mu }$ "restricted to" ${\displaystyle L_{x}}$ is the one-dimensional Lebesgue measure ${\displaystyle \lambda ^{1}}$, rather than the zero measure. The probability of a "two-dimensional" event ${\displaystyle E}$ could then be obtained as an integral of the one-dimensional probabilities of the vertical "slices" ${\displaystyle E\cap L_{x}}$: more formally, if ${\displaystyle \mu _{x}}$ denotes one-dimensional Lebesgue measure on ${\displaystyle L_{x}}$, then

$${\displaystyle \mu (E)=\int _{[0,1]}\mu _{x}(E\cap L_{x})\,\mathrm {d} x}$$

for any "nice" ${\displaystyle E\subseteq S}$. The disintegration theorem makes this argument rigorous in the context of measures on metric spaces.

(Hereafter, ${\displaystyle {\mathcal {P}}(X)}$ will denote the collection of Borel probability measures on a topological space ${\displaystyle (X,T)}$.) The assumptions of the theorem are as follows:

• Let ${\displaystyle Y}$ and ${\displaystyle X}$ be two Radon spaces (i.e. a topological space such that every Borel probability measure on it is inner regular, e.g. separably metrizable spaces; in particular, every probability measure on it is outright a Radon measure).
• Let ${\displaystyle \mu \in {\mathcal {P}}(Y)}$.
• Let ${\displaystyle \pi :Y\to X}$ be a Borel-measurable function. Here one should think of ${\displaystyle \pi }$ as a function to "disintegrate" ${\displaystyle Y}$, in the sense of partitioning ${\displaystyle Y}$ into ${\displaystyle \{\pi ^{-1}(x)\ |\ x\in X\}}$. For example, for the motivating example above, one can define ${\displaystyle \pi ((a,b))=a}$, ${\displaystyle (a,b)\in [0,1]\times [0,1]}$, which gives that ${\displaystyle \pi ^{-1}(a)=a\times [0,1]}$, a slice we want to capture.
• Let ${\displaystyle \nu \in {\mathcal {P}}(X)}$ be the pushforward measure ${\displaystyle \nu =\pi _{*}(\mu )=\mu \circ \pi ^{-1}}$. This measure provides the distribution of ${\displaystyle x}$ (which corresponds to the events ${\displaystyle \pi ^{-1}(x)}$).

The conclusion of the theorem: There exists a ${\displaystyle \nu }$-almost everywhere uniquely determined family of probability measures ${\displaystyle \{\mu _{x}\}_{x\in X}\subseteq {\mathcal {P}}(Y)}$, which provides a "disintegration" of ${\displaystyle \mu }$ into ${\displaystyle \{\mu _{x}\}_{x\in X}}$, such that:

• the function ${\displaystyle x\mapsto \mu _{x}}$ is Borel measurable, in the sense that ${\displaystyle x\mapsto \mu _{x}(B)}$ is a Borel-measurable function for each Borel-measurable set ${\displaystyle B\subseteq Y}$;
• ${\displaystyle \mu _{x}}$ "lives on" the fiber ${\displaystyle \pi ^{-1}(x)}$: for ${\displaystyle \nu }$-almost all ${\displaystyle x\in X}$,
  $${\displaystyle \mu _{x}\left(Y\setminus \pi ^{-1}(x)\right)=0,}$$

  
  and so ${\displaystyle \mu _{x}(E)=\mu _{x}(E\cap \pi ^{-1}(x))}$;
• for every Borel-measurable function ${\displaystyle f:Y\to [0,\infty ]}$,
  $${\displaystyle \int _{Y}f(y)\,\mathrm {d} \mu (y)=\int _{X}\int _{\pi ^{-1}(x)}f(y)\,\mathrm {d} \mu _{x}(y)\,\mathrm {d} \nu (x).}$$

  
  In particular, for any event ${\displaystyle E\subseteq Y}$, taking ${\displaystyle f}$ to be the indicator function of ${\displaystyle E}$,^[1]
  $${\displaystyle \mu (E)=\int _{X}\mu _{x}(E)\,\mathrm {d} \nu (x).}$$

  

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